Supervector spaces look a lot like the category of representations of $\mathbb{Z}/2\mathbb{Z}$ - the even part corresponds to the copies of the trivial representation and the odd part corresponds to the copies of the sign representation. This definition gives the usual morphisms, but it does not account for the braiding of the tensor product, which I've asked about previously.

Since the monoidal structure on $\text{Rep}(G)$ comes from the comultiplication on the group Hopf algebra, which in this case is $H = \mathbb{C}[\mathbb{Z}/2\mathbb{Z}]$, it seems reasonable to guess that the unusual tensor product structure comes from replacing the usual comultiplication, which is $\Delta g \mapsto g \otimes g$, with something else, or alternately from placing some extra structure on $H$. What extra structure accounts for the braiding?

3 Answers
3

The answer is yes, but comultiplication is not what you change. The symmetrizer (or braiding as you call it) is given by an $R$-element in $H \otimes H$ that makes $H$ into a triangular Hopf algebra. (The Wikipedia article says quasitriangular; triangular means that plus that the modified switching map $\widetilde{R}$ is an involution.)

The $R$ element is the correction to the usual switching map $x \otimes y \mapsto y \otimes x$. You can solve for it directly in terms of the usual description of the switching map on supervector spaces, and worry later about what axioms it satisfies. Let $1$ and $a$ be the group elements of $H$ and let $\epsilon$ and $\sigma$ be the dual vectors on $H$ which are the trivial and sign representations. Then
we want
$$(\epsilon \otimes \epsilon)(R) = 1 \quad (\epsilon \otimes \sigma)(R) = 1 \quad (\sigma \otimes \epsilon)(R) = 1 \quad (\sigma \otimes \sigma)(R) = -1,$$
because that is the correction in the modified switching map $x \otimes y \mapsto (-1)^{|x||y|} y \otimes x$. You can solve the linear system of equation to obtain
$$R = \frac{1 \otimes 1 + a \otimes 1 + 1 \otimes a - a \otimes a}{2}.$$

As a monoidal category, supervector spaces are the same as $\mathbb{Z}/2\mathbb{Z}$ representations, so you can think of them as representations of the Hopf algebra $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}]$.

However, the symmetric structure is different, so the thing you need to do is make $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}]$ into a quasi-triangular Hopf algebra in an interesting way, that is, change things so that the map from $V\otimes W\to W\otimes V$ isn't just flipping, it's flipping followed by an element of $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}]\otimes \mathbb{C}[\mathbb{Z}/2\mathbb{Z}]$ which is called the R-matrix. I'll leave actually writing this element as an exercise to the reader, but it's uniquely determined by acting by -1 on sign tensor sign and by 1 on everything else.

As an exercise for another poster in this case. Maybe you are right that an exercise for the reader would have been better.
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Greg KuperbergNov 8 '09 at 16:06

As a follow-up question, the Wikipedia article mentions something about quasi-invariants of knots. What invariant is obtained here?
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Qiaochu YuanNov 8 '09 at 16:17

1

The article should say invariant of braids rather than "quasi-invariant" of anything. The invariant in this case is almost trivial because of triangularity --- left crossing equals right crossing. For even strands it's trivial outright and for odd strands it's the sign representation of the braid group.
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Greg KuperbergNov 8 '09 at 16:22

Majid's <em>Foundations of quantum group theory</em> book explains this supersymmetry example of generating interesting monoidal category by nontrivial choice of R-matrix at length, as well as interesting "anyonic" generalizations of this example.
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Zoran SkodaFeb 25 '10 at 18:45

The correct Hopf algebra is not the group algebra kC_2 but its dual kC_2^*. If the characteristic of k is not 2 then accidentally kC_2\cong kC_2^*. Otherwise, Greg answered the question as your new R-element is no longer trivial 1\otimes 1 but 1 \otimes - 2\delta_x \otimes \delta_x where \delta_x is the delta function on the generator of C_2.